What are the solutions to the equation z^2-12z+36=-64 ?

The solutions to the equation z^2-12z+36=-64 are z=6+8i,z=6-8i

Find the zeros of z^2-12z+36=-64

The zeros of z^2-12z+36=-64 are z=6+8i,z=6-8i

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z^{2} - 12 z + 36 = - 64

z = 6 + 8 i, z = 6 - 8 i

Solution steps

z^{2} - 12 z + 36 = - 64

Move

64

to the left side

z^{2} - 12 z + 100 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 1 \cdot 100}{2 \cdot 1}

Simplify

(- 12)^{2} - 4 \cdot 1 \cdot 100 : 16 i

z_{1, 2} = \frac{- ( - 12 ) \pm 16 i}{2 \cdot 1}

Separate the solutions

z_{1} = \frac{- ( - 12 ) + 16 i}{2 \cdot 1}, z_{2} = \frac{- ( - 12 ) - 16 i}{2 \cdot 1}

z = \frac{- ( - 12 ) + 16 i}{2 \cdot 1} : 6 + 8 i

z = \frac{- ( - 12 ) - 16 i}{2 \cdot 1} : 6 - 8 i

The solutions to the quadratic equation are:

z = 6 + 8 i, z = 6 - 8 i

3 (x + 7)^{2} = 7

(x) = 3 x^{2} - 48 x + 350

(x - 6)^{2} + 10 (x - 6) = 0

(x + 13)^{2} = - 9

6 x^{2} - 80 x + 160 = 0

What are the solutions to the equation z^2-12z+36=-64 ?
The solutions to the equation z^2-12z+36=-64 are z=6+8i,z=6-8i
Find the zeros of z^2-12z+36=-64
The zeros of z^2-12z+36=-64 are z=6+8i,z=6-8i